X and Y are uniform random variables. X uniformly distributed on $[0, \frac{L}{2}]$ and Y is uniformly distibuted on $[\frac{L}{2}, L]$.
What is the probability that $X$ is located closer to $Y$ than to $0\,?$ $$$$ My attempt (which haven't got too far):
I am looking for the set of all points ${\{(x,y)\;|\;\;y-x<x,\;0\le x \le \frac{L}{2}, \; \frac{L}{2}\le y \le L \}}$ out of ${\{(x,y)\;|\;\;\;0\le x \le \frac{L}{2}, \; \frac{L}{2}\le y \le L \}}$. Thus:
$P(Y-X<X) = P(Y-2X<0)=$ Well i'm not sure how to continue from here.
What is the simpliest way of solving this?
The idea is correct but you need to complete the proof. Consider the set of all points $(X,Y)$ satisfying your conditions. All the points must satisfy the conditions $0\le X\le R$, $R\le Y\le 1$ where $R$ is $L/2$ a fixed number between $0, 1$. So the set of all points you consider is a rectangle $U$ of height $1-R$ and width $R$, area $A=R-R^2$.
The subset you are interested in is the set $S$ of points $(X,Y)$ such that $X>Y-X$ or $Y<2X$. So $S$ consists of all points from $U$ under the line $Y=2X$. The area $A_S$ of $S$ depends on $R$ and can be easily found. Indeed, let, say $R=2/3$. Then $S$ is a trapezium with height $1-R=1/3$, and bases $R-R/2=1/3$ and $R-1/2=1/6$, so $A_S=1/2(1/3+1/6)1/3=1/12$. Then the probability is $A_S/A$ (in the above case when $R=2/3$, the probability is $(1/12)/(2/3-4/9)=9/24=3/8$).